Abow and arrow hunter walks toward the origin along the positive x-axis, with unit speed. at time 0, he is at x=10. his arrow (of unit length) is aimed always toward a rabbit hopping with constant velocity √5 in the first quadrant along the line y = 2x, at time 0 it is at the origin.

a. write down the vector function for the arrow at time
b. the hunter shoots (and misses) when closest to the rabbit, when is that?

Step-by-step explanation:

Complete question:

Vector Functions and Parametric Equations

A bow-and-arrow hunter walks toward the origin along the positive x-axis, with unit speed; at time 0 he is at x = 10. His arrow (of unit length) is aimed always toward a rabbit hopping with constant velocity √5 in the first quadrant along the line y = 2x; at time 0 it is at the origin.

a) Write down the vector function A(t) for the arrow at time t.

b) The hunter shoots (and misses) when closest to the rabbit; when is that?

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hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

answer: it actually depends on which is the independent variable and which is the dependent variable. let us say we're looking at a linear equation. y=2x+3. we theoretically could switch the equation to be x=(y-3)/2 and it would be the same line. so either a causes b or it can be that b causes a. so technically the answer is false

step-by-step explanation: